All categories

1 year ago

What is the frequency of a photon of emr with a wavelength of 2.55x10-3m?

Physics

1 answer:

olga nikolaevna [1]1 year ago

3 0

Answer:

f = 1.18 x 10¹¹ Hz

Explanation:

The equation used to find frequency is:

f = c / w

In this form, "f" represents the frequency (Hz), "c" represents the speed of light (3.0 x 10⁸ m/s), and "w" represents the wavelength (m).

Since you have been given the value of the constant (c) and wavelength, you can substitute these values into the equation to find frequency.

f = c / w <---- Formula

f = (3.0 x 10⁸ m/s) / w <---- Plug 3.0 x 10⁸ in "c"

f = (3.0 x 10⁸ m/s) / (2.55 x 10⁻³ m) <---- Plug 2.55 x 10⁻³ in "w"

f = 1.18 x 10¹¹ Hz <---- Divide

You might be interested in

Question for the day!

Alex17521 [72]

Answer:

1. <--> A.

2. <--> C.

3. <--> D

4. <--> B.

explanation: i know my science!

3 0

2 years ago

Read 2 more answers

Please please help me please please help please please help me please

Kaylis [27]

Answer:

Electromagnetic waves do not require any medium to travel whereas mechanical waves must have a medium to propagate.

So, Basically, it is B I believe.

Hope It Helps!

6 0

2 years ago

7. If 8 million kg of water flows over Niagara Falls each second, calculate the power available at the bottom of the falls.

Alexxx [7]

Answer:

The power will be "3.92×10⁹ Watts". A further explanation is given below.

Explanation:

The given values as per the question,

Rate,

= 8 million kg

Distance,

= 50 m

Gravity,

= 9.8 m/s²

As we know,

The power will be:

⇒ $Power = Rate\times Distance\times Gravity$

On putting the values, we get

⇒ $= 8\times 10^6\times 50\times 9.8$

⇒ $=3.92\times 10^9 \ Watts$

7 0

2 years ago

Which of the following is an example of a material that allows electrons to easily move through it?

AVprozaik [17]

Conductors allow for charge transfer through the free movement of electrons

6 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

2 years ago